Question 1 of 8: Step response of an open-circuited line with a shunt tap

Question 1: Step response of an open-circuited line with a shunt tap (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 24 V step from a 100 $\Omega$ source launches a wave onto a lossless 50 $\Omega$ line; a 150 $\Omega$ resistor is bridged across the line half way along, and the far end is left open.

Find. The instant the wavefront reaches the tap, the voltage step that appears across the 150 $\Omega$ resistor at that instant, and the final DC voltage across it.

Approach. Launch the incident wave with the generator’s own divider, propagate it at $v_p$ to the tap, treat the tap as a junction whose impedance is the shunt resistor in parallel with the continuing line, and finally read the steady state off the DC circuit that a lossless line degenerates into.

1. Launch the incident wave. At $t = 0^{+}$ the generator sees only the line’s characteristic impedance, because no reflection has had time to return: $$V^{+} = E\,\frac{Z_0}{R_g + Z_0} = 24\times\frac{50}{100+50} = 8.00\ \text{V}$$ This 8 V step, and the current $V^{+}/Z_0 = 0.16\ \text{A}$ that accompanies it, travel toward the tap.
2. (i) Time of arrival at the tap. The line is lossless and non-dispersive, so the front travels at the stated phase velocity: $$t_1 = \frac{d_1}{v_p} = \frac{10\times10^{3}}{2\times10^{8}} = \boxed{50\ \mu\text{s}}$$
3. Impedance seen at the tap. This is the marked step. The resistor is connected across the line, not at its end, so the arriving wave sees the resistor and the second 10 km section in parallel. That section is 10 km of matched-looking line, which for the first transit presents exactly $Z_0$: $$Z_J = R \parallel Z_0 = \frac{150\times50}{150+50} = 37.5\ \Omega$$
4. Reflection and transmission at the tap. With the junction impedance known, $$\Gamma_J = \frac{Z_J - Z_0}{Z_J + Z_0} = \frac{37.5-50}{37.5+50} = -\frac{1}{7} = -0.1429$$ The node voltage is the incident wave plus its reflection, i.e. the transmission coefficient $\tau = 1 + \Gamma_J = 6/7$ acting on $V^{+}$.
5. (ii) Amplitude of the initial voltage. Substituting, $$V_{\text{init}} = V^{+}\left(1+\Gamma_J\right) = 8.00\times\frac{6}{7} = \boxed{6.86\ \text{V}}$$ The same 6.86 V step continues down the second section toward the open end, while a reflected $-1.14\ \text{V}$ step heads back to the generator.
6. How long that plateau lasts. The transmitted step reaches the open end after a further $d_2/v_p = 50\ \mu\text{s}$, reflects with $\Gamma = +1$, and returns to the tap at $t = 50 + 2(50) = 150\ \mu\text{s}$. The 6.86 V value is therefore the flat top observed for the first 100 $\mu\text{s}$ after arrival, which is what “initial voltage” means.
7. (iii) Final steady state. A lossless line carries no DC voltage drop, so once all reflections have died away the network is a plain resistive circuit. The open-circuited far section draws no direct current at all and therefore drops out entirely, leaving $E$, $R_g$ and $R$ in series: $$V_{\infty} = E\,\frac{R}{R_g + R} = 24\times\frac{150}{100+150} = \boxed{14.4\ \text{V}}$$

The three answers tell a consistent physical story: the resistor first sees only 6.86 V because the far section is still absorbing energy as though it were matched, and it then climbs in successive bounces to 14.4 V once the open end has sent everything back. The generator reflection coefficient here is $\Gamma_g = (100-50)/(100+50) = +1/3$, so the bounce sequence converges geometrically rather than ringing.